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Question 44 of 52

Q.If A + I₃ = (1 3 4; -1 1 3; -2 -3 1), find (A + I₃)(A - I₃). OR If 3A = (-1 2 -2; -2 1 2; 2 2 1), then show that A is an orthogonal matrix. Hence find A^-1.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2024Subjective· 4mImportance★★★★★
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First recover AA by subtracting I3I_3 from the given matrix, then compute A−I3A-I_3 and multiply the two matrices.

We are given A+I3=(134−113−2−31)A+I_3 = \begin{pmatrix}1&3&4\\-1&1&3\\-2&-3&1\end{pmatrix}.

Subtracting I3I_3 from both sides gives AA:

A=(134−113−2−31)−(100010001)=(034−103−2−30)A = \begin{pmatrix}1&3&4\\-1&1&3\\-2&-3&1\end{pmatrix} - \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} = \begin{pmatrix}0&3&4\\-1&0&3\\-2&-3&0\end{pmatrix}

Then A−I3=(034−103−2−30)−(100010001)=(−134−1−13−2−3−1)A-I_3 = \begin{pmatrix}0&3&4\\-1&0&3\\-2&-3&0\end{pmatrix} - \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix} = \begin{pmatrix}-1&3&4\\-1&-1&3\\-2&-3&-1\end{pmatrix}.

Now multiply P=A+I3P=A+I_3 by Q=A−I3Q=A-I_3 (row of PP times column of QQ):

Row 1 of PP is (1,3,4)(1,3,4): with columns of QQ, (−1,−1,−2)(-1,-1,-2), (3,−1,−3)(3,-1,-3), (4,3,−1)(4,3,-1):

1(−1)+3(−1)+4(−2)=−121(-1)+3(-1)+4(-2)=-12;  1(3)+3(−1)+4(−3)=−12\ 1(3)+3(-1)+4(-3)=-12;  1(4)+3(3)+4(−1)=9\ 1(4)+3(3)+4(-1)=9

Row 2 of PP is (−1,1,3)(-1,1,3):

−1(−1)+1(−1)+3(−2)=−6-1(-1)+1(-1)+3(-2)=-6;  −1(3)+1(−1)+3(−3)=−13\ -1(3)+1(-1)+3(-3)=-13;  −1(4)+1(3)+3(−1)=−4\ -1(4)+1(3)+3(-1)=-4

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